The given question is a medium difficulty 2-mark question from the chapter Arithmetic Progressions. Concept tested: Applying the formula to compute the nth term of an arithmetic progression. The question appeared in Section B of the 2016 CBSE class 10 board paper.

Question 9: The 4^{th} term of an A.P. is zero. Prove that the 25^{th} term of the A.P. is three times its 11^{th} term

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The n^{th}term of AP, a_{n}= a_{1}+ (n - 1) d

where a_{1}is the first term and d is the common difference

**Given Data**: a_{4} = 0

i.e., a_{4} = a_{1} + (4 - 1) d = 0

or a_{1} + 3d = 0

or a_{1} = -3d ..... (1)

a_{25} = a_{1} + (25 - 1 ) d = a_{1} + 24d ..... (2)

Substitute a_{1} = -3d in eqn (2)

a_{25} = -3d + 24d = 21d ..... (3)

a_{11} = a_{1} + (11 - 1) d = a_{1} + 10d ..... (4)

Substitute a_{1} = -3d in eqn (4)

a_{11} = -3d + 10d = 7d ..... (5)

Compare equations (3) and (5)

Equation (3): a_{25} = 21d

Equation (5): a_{11} = 7d

21d = 3 × 7d

Or a_{25} = 3 × a_{11}

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